MATH 131
Final Exam
June 29,2023
8. (10 points) A manufacturer of a particular product makes $x$ bats at a cost of $C(x)=8 x+20$ dollars. The revenue from the sale of the product is given by $R(x)=100 x-0.5 x^{2}$ dollars.
i. ( 6 points) How many products should be manufactured and sold to maximize profit?
ii. (4 points) Use the first or second derivative to show that the number of products you found gives a maximum profit

Step 1 ：Given the cost function \(C(x) = 8x + 20\) and the revenue function \(R(x) = 100x - 0.5x^2\), the profit function \(P(x)\) is given by \(P(x) = R(x) - C(x)\).

Step 2 ：Substituting the given functions into the profit function, we get \(P(x) = -0.5x^2 + 92x - 20\).

Step 3 ：To find the maximum of the profit function, we take the derivative of the function and set it equal to zero. This gives us \(P'(x) = 92 - x\).

Step 4 ：Solving \(P'(x) = 0\) gives us the critical point \(x = 92\).

Step 5 ：We then use the second derivative test to determine whether this critical point is a maximum. The second derivative of the profit function is \(P''(x) = -1\).

Step 6 ：Since \(P''(x)\) is negative, this confirms that \(x = 92\) is a maximum.

Step 7 ：\(\boxed{\text{i. The manufacturer should produce and sell 92 products to maximize profit.}}\)

Step 8 ：\(\boxed{\text{ii. The second derivative of the profit function is negative at x=92, which confirms that this point is a maximum.}}\)